KCET · Maths · Determinants
If \(A=\left|\begin{array}{ll}x & 1 \\ 1 & x\end{array}\right|\) and \(B=\left|\begin{array}{lll}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{array}\right|\), then \(\frac{d B}{d x}\) is
- A \(3 A\)
- B \(-3 B\)
- C \(3 B+1\)
- D \(1-3 A\)
Answer & Solution
Correct Answer
(A) \(3 A\)
Step-by-step Solution
Detailed explanation
\(B=\left|\begin{array}{lll}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{array}\right|\)
\(=\left|\begin{array}{lll}1 & 1 & 1 \\ 0 & x & 1 \\ 0 & 1 & x\end{array}\right|+\left|\begin{array}{lll}x & 0 & 1 \\ 1 & 1 & 1 \\ 1 & 0 & x\end{array}\right|+\left|\begin{array}{lll}x & 1 & 0 \\ 1 & x & 0 \\ 1 & 1 & 1\end{array}\right|\)
\(=\left(x^2-1\right)+\left(x^2-1\right)+\left(x^2-1\right)\)
\(=3\left(x^2-1\right)\)
\(=3 A \quad\left[\because A=\left|\begin{array}{ll}x & 1 \\ 1 & x\end{array}\right|=x^2-1\right]\)
\(=\left|\begin{array}{lll}1 & 1 & 1 \\ 0 & x & 1 \\ 0 & 1 & x\end{array}\right|+\left|\begin{array}{lll}x & 0 & 1 \\ 1 & 1 & 1 \\ 1 & 0 & x\end{array}\right|+\left|\begin{array}{lll}x & 1 & 0 \\ 1 & x & 0 \\ 1 & 1 & 1\end{array}\right|\)
\(=\left(x^2-1\right)+\left(x^2-1\right)+\left(x^2-1\right)\)
\(=3\left(x^2-1\right)\)
\(=3 A \quad\left[\because A=\left|\begin{array}{ll}x & 1 \\ 1 & x\end{array}\right|=x^2-1\right]\)
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